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Ask QuestionPosted by Sushant Patil 6 years ago
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Sia ? 6 years ago
LHS = {tex}\frac { 1 + \sin \theta - \cos \theta } { 1 + \sin \theta + \cos \theta }{/tex}
= {tex}\frac { ( 1 - \cos \theta ) + \sin \theta } { ( 1 + \cos \theta ) + \sin \theta }{/tex} = {tex}\frac { 2 \sin ^ { 2 } \frac { \theta } { 2 } + 2 \sin \frac { \theta } { 2 } \cdot \cos \frac { \theta } { 2 } } { 2 \cos ^ { 2 } \frac { \theta } { 2 } + 2 \sin \frac { \theta } { 2 } \cdot \cos \frac { \theta } { 2 } }{/tex}
[{tex}\because{/tex} sin2x = {tex}\frac { 1 - \cos 2 x } { 2 } \Rightarrow{/tex} {tex}2 sin^2x = 1 - cos 2x{/tex} and {tex}2sin^2 \frac{x}{2} = 1 - cosx{/tex} and {tex}2 cos^2{/tex} {tex}\frac { x } { 2 }{/tex} {tex}= 1 + cosx{/tex} and {tex}sinx = 2sin{/tex} {tex}\frac { x } { 2 }{/tex} {tex}\times{/tex} {tex}cos{/tex} {tex}\frac { x } { 2 }{/tex}]
= {tex}\frac { \sin ^ { 2 } \frac { \theta } { 2 } + \sin \frac { \theta } { 2 } \cdot \cos \frac { \theta } { 2 } } { \cos ^ { 2 } \frac { \theta } { 2 } + \sin \frac { \theta } { 2 } \cdot \cos \frac { \theta } { 2 } } = \frac { \sin \frac { \theta } { 2 } \left[ \sin \frac { \theta } { 2 } + \cos \frac { \theta } { 2 } \right] } { \cos \frac { \theta } { 2 } \left[ \cos \frac { \theta } { 2 } + \sin \frac { \theta } { 2 } \right] }{/tex}
= {tex}\frac { \sin \frac { \theta } { 2 } } { \cos \frac { \theta } { 2 } }{/tex} = tan {tex}\frac { \theta } { 2 }{/tex} = RHS
{tex}\therefore{/tex} LHS = RHS
Hence proved.
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